Complete Set of Commuting Operators

There is only one way of doing this. It is very direct, and it consists of exhibiting another ``matrix'', i.e., operator, which

1. has the same domain as $\nabla^2$ ,
2. has the same eigenvectors that $\nabla^2$ has, but
3. has eigenvalues which are different for different eigenvectors.

Examples of such ``matrices'' are

$$\frac{1}{i}~\frac{\partial}{\partial x} \equiv P_x~~\quad~~\textrm{and} ~~\quad~~ \frac{1}{i}~\frac{\partial}{\partial y} \equiv P_y\,.$$

Their eigenvectors are the plane wave solutions,

$$P_xe^{i\vec k\cdot\vec x} = k_x e^{i\vec k\cdot\vec x}~\quad~\textrm{and} ~\quad~P_ye^{i\vec k\cdot\vec x} = k_ye^{i\vec k\cdot\vec x}\,,$$

a fact which is also the case for the Helmholtz operator,

$$-\nabla^2 e^{i\vec k\cdot\vec x} = k^2\cdot e^{i\vec k\cdot\vec x}\,.$$

However, note that the eigenvalues, $k_x$ and $k_y$ , are different for different plane wave solutions. Thus one has available a very succinct way of characterizing the elements of each degenerate subspace for each eigenvalue $k^2$ of $-\nabla^2$ . This way consists of the statement that the eigenbasis spanning this subspace be labelled by the eigenvalue triplet

$$(k_x,k_y,k^2)$$

of the corresponding three operators

$$\{ P_x,P_y,-\nabla^2\}\,.$$

This labelling is unique, i.e., the correspondence

$$\{ (k_x,k_y,k^2)\}\leftrightarrow\{ e^{i(k_xx+k_yy)} = \psi_{k_x,k_y,k^2} (x,y)\}$$

is unique.The operators $\{ P_x,P_y,\nabla^2\}$ form what is called a complete set of commuting operators because their eigenvalues $(k_x,k_y,k^2)$ serve as sufficient labels to uniquely identify each of their (common) eigenbasis elements for the vector space of solutions to the Hermholtz equation. No additional labels are necessary.

In addition, notice the following: that the three operators $(P_x,P_y,- \nabla^2 \equiv P^2_x+P^2_y)$ have the same eigenvectors implies that they commute

$$[P_x,P_y] = [P_x,\nabla^2] = [P_y,\nabla^2] = 0\,.$$

In fact, one can show that two operators, each having an eigenbasis for the vector space, commute if and only if they have an eigenbasis in common. This commutativity is a representation independent way of stating that the invariant subspaces of one operator coincide with the invariant subspaces of the other operator, even though their eigenvalues do not. An alternate way of saying this is that

$$\textrm{span} \{ e^{i(k_x x+k_yy)}\colon k^2_x+k^2_y = k^2\}$$

is a subspace invariant under $P_x,P_y$ , and $\nabla^2$ . To illustrate the commonality of these subspaces, consider the one-dimensional subspace spanned by the eigenvector of the nondegenerate eigenvalue $k_x$ of $P_x$ ,

$$P_x\psi =k_x \psi.$$

What can one say about $\nabla^2 \psi$ ? To find out consider $P_x\nabla^2 \psi$ . Using the fact that $P_x\nabla^2 =\nabla^2P_x$ one has

$$P_x\nabla^2 \psi=\nabla^2P_x \psi =k_x \nabla^2 \psi~.$$

The fact that the eigenvalue $k_x$ is nondegenerate implies that $\nabla^2 \psi$ is a multiple of $\psi$ :

$$\nabla^2 \psi =\lambda \psi~.$$

Thus $\psi$ is also an eigenvector of $\nabla^2$ . Thus we have proved an important Theorem

Suppose that

$$[P_x,\nabla^2]=0$$

and $\psi$ is an eigenvector belonging to the nondegenerate eigenvalue $k_x$ :

$$P_x\psi =k_x \psi;$$

then $\psi$ is also an eigenvector of $\nabla^2$ .